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Pasch’s theorem

Theorem.

(Pasch) Let △⁢a⁢b⁢cnormal-△abc\triangle abc be a triangle with
non-collinear vertices a,b,cabca,b,c in a linear ordered geometry.
Suppose a line ℓnormal-ℓ\ell intersects one side, say open line segment a⁢b¯normal-¯ab\overline{ab}, at a
point strictly between aaa and bbb, then ℓnormal-ℓ\ell also intersects exactly one of the following:

Proof.

First, note that verticesaaa and bbb are on opposite sides of lineℓnormal-ℓ\ell. Then either ccc lies on ℓnormal-ℓ\ell, or ccc does not. if ccc does not, then it must lie on either side (half plane) of ℓnormal-ℓ\ell. In other words, ccc and aaa must be on the opposite sides of ℓnormal-ℓ\ell, or ccc and bbb must be on the opposite sides of ℓnormal-ℓ\ell.
If ccc and aaa are on the opposite sides, ℓnormal-ℓ\ell has a non-empty intersection with a⁢c¯normal-¯ac\overline{ac}. But if ccc and aaa are on the opposite sides, then ccc and bbb are on the same side, which means that b⁢c¯normal-¯bc\overline{bc} does not intersectℓnormal-ℓ\ell.
∎

Remark
A companion propertystates that if line ℓnormal-ℓ\ellpasses through one vertex aaa of a triangle△⁢a⁢b⁢cnormal-△abc\triangle abc and at least one other point on △⁢a⁢b⁢cnormal-△abc\triangle abc, then it must intersect exactly one of the following: